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My question is how to calculate

$$ \int_{\mathbb{R}^n} f(x) \delta(h(\mathbf{x})) \delta(g(\mathbf{x})) \mathrm{d} \mathbf{x} $$

(In my case both $h(\mathbf{x})$ and $g(\mathbf{x})$ are linear function, it would be nice to have a more general solution)

I learn from Wikipedia that $$ \int_{\mathbb{R}^n} f(\mathbf{x}) \delta(g(\mathbf{x})) \mathrm{d} \mathbf{x}=\int_{g^{-1}(0)} \frac{f(\mathbf{x})}{|\nabla g|} \mathrm{d} \sigma(\mathbf{x}) $$

So I think $$ \int_{\mathbb{R}^n} f(x) \delta(h(\mathbf{x})) \delta(g(\mathbf{x})) \mathrm{d} \mathbf{x} = \int_{g^{-1}(0)} \frac{f(\mathbf{x})\delta(h(\mathbf{x}))}{|\nabla g|} \mathrm{d} \sigma(\mathbf{x}) $$

But I don't know what should I do in next step, if I reapet the same method I get

$$ \int_{g^{-1}(0)} \frac{f(\mathbf{x})\delta(h(\mathbf{x}))}{|\nabla g|} \mathrm{d} \sigma(\mathbf{x}) = \int_{g^{-1}(0), h^{-1}(0)} \frac{f(\mathbf{x})}{|\nabla g||\nabla h|} \mathrm{d} s(\mathbf{x}) $$

If it is correct, I don't know if $\nabla h$ is calculated in original $\mathbb{R}^n$ space or $g^{-1}(0)$ space?

Stupid Boy
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  • Multiplication of two Dirac deltas is not defined in the standard theory of generalized functions. – Saleh Nov 08 '22 at 08:26
  • @Saleh True, but distributions on hypersurfaces are well-defined. $\delta(g) \delta(h)$ usually means a distribution supported on $g = h = 0$. If we require the sifting property of the delta function to hold and the change of variables rule to hold, we obtain a definition of $\delta(g) \delta(h)$ which is independent of how we parametrize $g = 0$ and $h = 0$ (assuming that there exists a coordinate system in which $g$ and $h$ are coordinate surfaces). We'll get this formula, where $\nabla$ is the usual $\mathbb R^n$ gradient. – Maxim Nov 12 '22 at 13:44

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